## Lecture 13: Zero-sum games using Experts

The Oracle. My explanation for why the oracle was an easy problem was not very clear. Let me try again. Let us define

$\displaystyle K = \{ x \in {\mathbb R}^n \mid x \geq 0, c^\intercal x = OPT \}.$

Notice that this has ${n}$ non-negativity constraints, and one equality constraint. Very simple constraints. Now the oracle asks: given ${\alpha \in {\mathbb R}^n, \beta \in {\mathbb R}}$, find me a point ${x \in K}$ such that

$\displaystyle \alpha^\intercal x \geq \beta,$

or else tell me no such point exists. How would you solve this?

Let’s just try to maximize ${\alpha^\intercal x}$ for ${x \in K}$. If we can make ${\alpha^\intercal x}$ at least ${\beta}$, then we can return that point. If we cannot reach ${\beta}$, we return “No such point.”

OK, suppose all entries of ${c}$ are positive. Then I claim you might as well make ${x}$ non-zero in a single coordinate. Which coordinate? The entry ${j}$ that maximizes

$\displaystyle \frac{OPT}{c_j} \cdot \alpha_j.$

Then you’ll make ${x_j = \frac{OPT}{c_j}}$ to satisfy ${x \in K}$. (Of course, if ${\frac{OPT}{c_j} \cdot \alpha_j < \beta}$, you’ll return “No such point.”)

If ${c}$ has negative entries, then things are a more interesting. You can still ensure that ${x}$ will be non-zero in at most two entries. And then you have to do a simple case analysis. I’ll omit the details here; let me know if you’d like me to say more.

The next time, we’ll show that if you can solve the oracle problem you can solve LPs (up to ${\varepsilon}$ additive error).